Proof verification of the language of all palindromes as being context-free












0












$begingroup$


Consider that the language L of all palindromes over $Sigma = {0,1}^*$ is not context-free. The following is my attempt at a proof by contradiction.



I am new to proof writing and I am wondering if the proof is correct, and if it proceeds in a connected logical sequence. I think I have all the cases covered, but I am not too sure.



enter image description here






proof-verification proof-writing context-free-grammar pumping-lemma palindrome







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Could you explain why $uwy = 0^k1^{2k}o^k$? If I set $w = 1, v = x = 1$, then neither $x$ nor $v$ equals $epsilon$, $|vwx| < p$ and $uv^iwx^iy in L$, but $uwy$ is not of the form you state. Apart from that, there's a very simple grammar for palindromes: $w := "" | "1" | "0" | "1" w "1" | "0" w "0"$ that looks pretty context free to me
    $endgroup$
    – Ronald
    Dec 6 '18 at 0:33










  • $begingroup$
    @Ronald could you elaborate? Are you saying that the language of palindromes is context free?
    $endgroup$
    – SeesSound
    Dec 6 '18 at 1:02


















0












$begingroup$


Consider that the language L of all palindromes over $Sigma = {0,1}^*$ is not context-free. The following is my attempt at a proof by contradiction.



I am new to proof writing and I am wondering if the proof is correct, and if it proceeds in a connected logical sequence. I think I have all the cases covered, but I am not too sure.



enter image description here






proof-verification proof-writing context-free-grammar pumping-lemma palindrome







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Could you explain why $uwy = 0^k1^{2k}o^k$? If I set $w = 1, v = x = 1$, then neither $x$ nor $v$ equals $epsilon$, $|vwx| < p$ and $uv^iwx^iy in L$, but $uwy$ is not of the form you state. Apart from that, there's a very simple grammar for palindromes: $w := "" | "1" | "0" | "1" w "1" | "0" w "0"$ that looks pretty context free to me
    $endgroup$
    – Ronald
    Dec 6 '18 at 0:33










  • $begingroup$
    @Ronald could you elaborate? Are you saying that the language of palindromes is context free?
    $endgroup$
    – SeesSound
    Dec 6 '18 at 1:02
















0












0








0





$begingroup$


Consider that the language L of all palindromes over $Sigma = {0,1}^*$ is not context-free. The following is my attempt at a proof by contradiction.



I am new to proof writing and I am wondering if the proof is correct, and if it proceeds in a connected logical sequence. I think I have all the cases covered, but I am not too sure.



enter image description here






proof-verification proof-writing context-free-grammar pumping-lemma palindrome







share|cite|improve this question









$endgroup$




Consider that the language L of all palindromes over $Sigma = {0,1}^*$ is not context-free. The following is my attempt at a proof by contradiction.



I am new to proof writing and I am wondering if the proof is correct, and if it proceeds in a connected logical sequence. I think I have all the cases covered, but I am not too sure.



enter image description here






proof-verification proof-writing context-free-grammar pumping-lemma palindrome




proof-verification proof-writing context-free-grammar pumping-lemma palindrome






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 5 '18 at 23:57









SeesSoundSeesSound

858




858








  • 3




    $begingroup$
    Could you explain why $uwy = 0^k1^{2k}o^k$? If I set $w = 1, v = x = 1$, then neither $x$ nor $v$ equals $epsilon$, $|vwx| < p$ and $uv^iwx^iy in L$, but $uwy$ is not of the form you state. Apart from that, there's a very simple grammar for palindromes: $w := "" | "1" | "0" | "1" w "1" | "0" w "0"$ that looks pretty context free to me
    $endgroup$
    – Ronald
    Dec 6 '18 at 0:33










  • $begingroup$
    @Ronald could you elaborate? Are you saying that the language of palindromes is context free?
    $endgroup$
    – SeesSound
    Dec 6 '18 at 1:02
















  • 3




    $begingroup$
    Could you explain why $uwy = 0^k1^{2k}o^k$? If I set $w = 1, v = x = 1$, then neither $x$ nor $v$ equals $epsilon$, $|vwx| < p$ and $uv^iwx^iy in L$, but $uwy$ is not of the form you state. Apart from that, there's a very simple grammar for palindromes: $w := "" | "1" | "0" | "1" w "1" | "0" w "0"$ that looks pretty context free to me
    $endgroup$
    – Ronald
    Dec 6 '18 at 0:33










  • $begingroup$
    @Ronald could you elaborate? Are you saying that the language of palindromes is context free?
    $endgroup$
    – SeesSound
    Dec 6 '18 at 1:02










3




3




$begingroup$
Could you explain why $uwy = 0^k1^{2k}o^k$? If I set $w = 1, v = x = 1$, then neither $x$ nor $v$ equals $epsilon$, $|vwx| < p$ and $uv^iwx^iy in L$, but $uwy$ is not of the form you state. Apart from that, there's a very simple grammar for palindromes: $w := "" | "1" | "0" | "1" w "1" | "0" w "0"$ that looks pretty context free to me
$endgroup$
– Ronald
Dec 6 '18 at 0:33




$begingroup$
Could you explain why $uwy = 0^k1^{2k}o^k$? If I set $w = 1, v = x = 1$, then neither $x$ nor $v$ equals $epsilon$, $|vwx| < p$ and $uv^iwx^iy in L$, but $uwy$ is not of the form you state. Apart from that, there's a very simple grammar for palindromes: $w := "" | "1" | "0" | "1" w "1" | "0" w "0"$ that looks pretty context free to me
$endgroup$
– Ronald
Dec 6 '18 at 0:33












$begingroup$
@Ronald could you elaborate? Are you saying that the language of palindromes is context free?
$endgroup$
– SeesSound
Dec 6 '18 at 1:02






$begingroup$
@Ronald could you elaborate? Are you saying that the language of palindromes is context free?
$endgroup$
– SeesSound
Dec 6 '18 at 1:02












1 Answer
1






active

oldest

votes


















0












$begingroup$

If a context free grammar exists that produces the language, then the language itself is context free.



A context free grammar is a 4-tuple $G = (V, Sigma, R, S)$ with the following properties:





  1. $V$ is a finite set; $v in V$ is called a non-terminal symbol


  2. $Sigma$ is a finite set of terminal symbols; $V cap Sigma =emptyset$


  3. $R$ is a finite relation $V rightarrow (V cup Sigma)^*$


  4. $S in V$ is the start symbol


With $V = {w}$, $Sigma = {0, 1}$, $S = w$ and the following map $R$:



$w rightarrow epsilon$
$w rightarrow 0$
$w rightarrow 1$
$w rightarrow 0w0$
$w rightarrow 1w1$



we've defined a context free grammar.



Note: if the empty string isn't considered to be a palindrome, the first production can be replaced by



$w rightarrow 00$
$w rightarrow 11$



Palindromes are, casually speaking, a special case of "well-formed parenthesis" (each opening parenthesis has a corresponding closing parenthesis) which is also known to be context free.



Context free languages can't "count" very well, but it isn't a problem to have an arbitrary number of matching pairs as this can be tested using a stack.
But the famous language $a^nb^nc^n$ isn't context free because there's no way to count to $n$.



Edit:

Strictly speaking I'll have to prove that the above grammar indeed produces the set of all palindromes.



First of all, if $w in L$, which means that $w$ is a palindrome, then also $0w0$ and $1w1$ are palindromes and therefore member of $L$.
Obviously $0, 1$ and $epsilon$ are palindromes.

In other words, the production rules only produce palindromes.



Now assume $p$ is a palindrome. Then $p = s_1s_2...s_nms_ns_{n-1}...s_1$, with $s_i in {0, 1}, i in {1, ..., n}$ and $m in {epsilon, 0, 1}$.



In order to produce this palindrome, we start with $w = m$ and then consecutively add the symbols $s_n, s_{n-1}, ..., s_1$ on either side of $w$, which is allowed by the last two production rules.



This proves that $p$ can be produced by the grammar above.

Together with the fact that $G$ produces palindromes only, this proves that $G$ is a valid grammar for $L$.






share|cite|improve this answer











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3027846%2fproof-verification-of-the-language-of-all-palindromes-as-being-context-free%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    If a context free grammar exists that produces the language, then the language itself is context free.



    A context free grammar is a 4-tuple $G = (V, Sigma, R, S)$ with the following properties:





    1. $V$ is a finite set; $v in V$ is called a non-terminal symbol


    2. $Sigma$ is a finite set of terminal symbols; $V cap Sigma =emptyset$


    3. $R$ is a finite relation $V rightarrow (V cup Sigma)^*$


    4. $S in V$ is the start symbol


    With $V = {w}$, $Sigma = {0, 1}$, $S = w$ and the following map $R$:



    $w rightarrow epsilon$
    $w rightarrow 0$
    $w rightarrow 1$
    $w rightarrow 0w0$
    $w rightarrow 1w1$



    we've defined a context free grammar.



    Note: if the empty string isn't considered to be a palindrome, the first production can be replaced by



    $w rightarrow 00$
    $w rightarrow 11$



    Palindromes are, casually speaking, a special case of "well-formed parenthesis" (each opening parenthesis has a corresponding closing parenthesis) which is also known to be context free.



    Context free languages can't "count" very well, but it isn't a problem to have an arbitrary number of matching pairs as this can be tested using a stack.
    But the famous language $a^nb^nc^n$ isn't context free because there's no way to count to $n$.



    Edit:

    Strictly speaking I'll have to prove that the above grammar indeed produces the set of all palindromes.



    First of all, if $w in L$, which means that $w$ is a palindrome, then also $0w0$ and $1w1$ are palindromes and therefore member of $L$.
    Obviously $0, 1$ and $epsilon$ are palindromes.

    In other words, the production rules only produce palindromes.



    Now assume $p$ is a palindrome. Then $p = s_1s_2...s_nms_ns_{n-1}...s_1$, with $s_i in {0, 1}, i in {1, ..., n}$ and $m in {epsilon, 0, 1}$.



    In order to produce this palindrome, we start with $w = m$ and then consecutively add the symbols $s_n, s_{n-1}, ..., s_1$ on either side of $w$, which is allowed by the last two production rules.



    This proves that $p$ can be produced by the grammar above.

    Together with the fact that $G$ produces palindromes only, this proves that $G$ is a valid grammar for $L$.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      If a context free grammar exists that produces the language, then the language itself is context free.



      A context free grammar is a 4-tuple $G = (V, Sigma, R, S)$ with the following properties:





      1. $V$ is a finite set; $v in V$ is called a non-terminal symbol


      2. $Sigma$ is a finite set of terminal symbols; $V cap Sigma =emptyset$


      3. $R$ is a finite relation $V rightarrow (V cup Sigma)^*$


      4. $S in V$ is the start symbol


      With $V = {w}$, $Sigma = {0, 1}$, $S = w$ and the following map $R$:



      $w rightarrow epsilon$
      $w rightarrow 0$
      $w rightarrow 1$
      $w rightarrow 0w0$
      $w rightarrow 1w1$



      we've defined a context free grammar.



      Note: if the empty string isn't considered to be a palindrome, the first production can be replaced by



      $w rightarrow 00$
      $w rightarrow 11$



      Palindromes are, casually speaking, a special case of "well-formed parenthesis" (each opening parenthesis has a corresponding closing parenthesis) which is also known to be context free.



      Context free languages can't "count" very well, but it isn't a problem to have an arbitrary number of matching pairs as this can be tested using a stack.
      But the famous language $a^nb^nc^n$ isn't context free because there's no way to count to $n$.



      Edit:

      Strictly speaking I'll have to prove that the above grammar indeed produces the set of all palindromes.



      First of all, if $w in L$, which means that $w$ is a palindrome, then also $0w0$ and $1w1$ are palindromes and therefore member of $L$.
      Obviously $0, 1$ and $epsilon$ are palindromes.

      In other words, the production rules only produce palindromes.



      Now assume $p$ is a palindrome. Then $p = s_1s_2...s_nms_ns_{n-1}...s_1$, with $s_i in {0, 1}, i in {1, ..., n}$ and $m in {epsilon, 0, 1}$.



      In order to produce this palindrome, we start with $w = m$ and then consecutively add the symbols $s_n, s_{n-1}, ..., s_1$ on either side of $w$, which is allowed by the last two production rules.



      This proves that $p$ can be produced by the grammar above.

      Together with the fact that $G$ produces palindromes only, this proves that $G$ is a valid grammar for $L$.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        If a context free grammar exists that produces the language, then the language itself is context free.



        A context free grammar is a 4-tuple $G = (V, Sigma, R, S)$ with the following properties:





        1. $V$ is a finite set; $v in V$ is called a non-terminal symbol


        2. $Sigma$ is a finite set of terminal symbols; $V cap Sigma =emptyset$


        3. $R$ is a finite relation $V rightarrow (V cup Sigma)^*$


        4. $S in V$ is the start symbol


        With $V = {w}$, $Sigma = {0, 1}$, $S = w$ and the following map $R$:



        $w rightarrow epsilon$
        $w rightarrow 0$
        $w rightarrow 1$
        $w rightarrow 0w0$
        $w rightarrow 1w1$



        we've defined a context free grammar.



        Note: if the empty string isn't considered to be a palindrome, the first production can be replaced by



        $w rightarrow 00$
        $w rightarrow 11$



        Palindromes are, casually speaking, a special case of "well-formed parenthesis" (each opening parenthesis has a corresponding closing parenthesis) which is also known to be context free.



        Context free languages can't "count" very well, but it isn't a problem to have an arbitrary number of matching pairs as this can be tested using a stack.
        But the famous language $a^nb^nc^n$ isn't context free because there's no way to count to $n$.



        Edit:

        Strictly speaking I'll have to prove that the above grammar indeed produces the set of all palindromes.



        First of all, if $w in L$, which means that $w$ is a palindrome, then also $0w0$ and $1w1$ are palindromes and therefore member of $L$.
        Obviously $0, 1$ and $epsilon$ are palindromes.

        In other words, the production rules only produce palindromes.



        Now assume $p$ is a palindrome. Then $p = s_1s_2...s_nms_ns_{n-1}...s_1$, with $s_i in {0, 1}, i in {1, ..., n}$ and $m in {epsilon, 0, 1}$.



        In order to produce this palindrome, we start with $w = m$ and then consecutively add the symbols $s_n, s_{n-1}, ..., s_1$ on either side of $w$, which is allowed by the last two production rules.



        This proves that $p$ can be produced by the grammar above.

        Together with the fact that $G$ produces palindromes only, this proves that $G$ is a valid grammar for $L$.






        share|cite|improve this answer











        $endgroup$



        If a context free grammar exists that produces the language, then the language itself is context free.



        A context free grammar is a 4-tuple $G = (V, Sigma, R, S)$ with the following properties:





        1. $V$ is a finite set; $v in V$ is called a non-terminal symbol


        2. $Sigma$ is a finite set of terminal symbols; $V cap Sigma =emptyset$


        3. $R$ is a finite relation $V rightarrow (V cup Sigma)^*$


        4. $S in V$ is the start symbol


        With $V = {w}$, $Sigma = {0, 1}$, $S = w$ and the following map $R$:



        $w rightarrow epsilon$
        $w rightarrow 0$
        $w rightarrow 1$
        $w rightarrow 0w0$
        $w rightarrow 1w1$



        we've defined a context free grammar.



        Note: if the empty string isn't considered to be a palindrome, the first production can be replaced by



        $w rightarrow 00$
        $w rightarrow 11$



        Palindromes are, casually speaking, a special case of "well-formed parenthesis" (each opening parenthesis has a corresponding closing parenthesis) which is also known to be context free.



        Context free languages can't "count" very well, but it isn't a problem to have an arbitrary number of matching pairs as this can be tested using a stack.
        But the famous language $a^nb^nc^n$ isn't context free because there's no way to count to $n$.



        Edit:

        Strictly speaking I'll have to prove that the above grammar indeed produces the set of all palindromes.



        First of all, if $w in L$, which means that $w$ is a palindrome, then also $0w0$ and $1w1$ are palindromes and therefore member of $L$.
        Obviously $0, 1$ and $epsilon$ are palindromes.

        In other words, the production rules only produce palindromes.



        Now assume $p$ is a palindrome. Then $p = s_1s_2...s_nms_ns_{n-1}...s_1$, with $s_i in {0, 1}, i in {1, ..., n}$ and $m in {epsilon, 0, 1}$.



        In order to produce this palindrome, we start with $w = m$ and then consecutively add the symbols $s_n, s_{n-1}, ..., s_1$ on either side of $w$, which is allowed by the last two production rules.



        This proves that $p$ can be produced by the grammar above.

        Together with the fact that $G$ produces palindromes only, this proves that $G$ is a valid grammar for $L$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 6 '18 at 14:02

























        answered Dec 6 '18 at 9:08









        RonaldRonald

        673510




        673510






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3027846%2fproof-verification-of-the-language-of-all-palindromes-as-being-context-free%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Berounka

            Image
            Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
            Read more

            Sphinx de Gizeh

            Image
            Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
            Read more

            Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

            Image
            3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
            Read more